The function is defined if...

**utsav** · September 24th 2009, 09:39 PM

(e^( cos^(-1) (log x^2) ))^1/2

The base of log is 4

**Prove It** · September 24th 2009, 09:41 PM

Originally Posted by utsav

(e^( cos^(-1) (log x^2) ))^1/2

The base of log is 4

What are you trying to do?

**utsav** · September 24th 2009, 09:44 PM

Originally Posted by Prove It

What are you trying to do?

The question is: The function is defined if..., we have to find the domain of the given function.

**Prove It** · September 24th 2009, 09:59 PM

Originally Posted by utsav

(e^( cos^(-1) (log x^2) ))^1/2

The base of log is 4

$f(x) = \sqrt{e^{\arccos{[\log{(x^2)}]}}}$ .

Note that the logarithm is only defined for $x > 0$ .

Since $x^2 \geq 0$ , we can already see that we can not let $x = 0$ .

Also note that $\arccos{x}$ is only defined for $-1 \leq x \leq 1$ .

The exponential function is always $>0$ , so there are not any further restrictions for the square root.

Thus the domain is $x \in [-1, 0)\cup(0, 1]$ .

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